#)Giải :
#)Giải :
Vì \(\widehat{AOC}\)và \(\widehat{BOD}\)là hai góc đối đỉnh \(\Rightarrow\widehat{AOC}=\widehat{BOD}\left(=70^o\right)\)
Vì \(\widehat{AOC}\)và \(\widehat{BOC}\)là hai góc kề bù
\(\Rightarrow\widehat{BOC}=180^o-\widehat{AOC}\)
\(=180^o-70^o\)
\(=110^o\)
\(\Rightarrow\widehat{BOC}=110^o\)
Vì \(\widehat{BOC}\)và \(\widehat{AOD}\)là hai góc đối đỉnh \(\Rightarrow\widehat{BOC}=\widehat{AOD}\left(=110^o\right)\)
#~Will~be~Pens~#
Theo đề bài biết :
\(\widehat{AOC}\)- \(\widehat{BOC}\)= 70o
Ngoài ra còn biết :
\(\widehat{AOC}\)+ \(\widehat{BOC}\)= 180o ( kề bù )
\(\rightarrow\)\(\widehat{AOC}\)= ( 70o + 180o ) : 2 = 125o
\(\rightarrow\)\(\widehat{BOC}\)= 180o - 125o = 55o
Có \(\widehat{AOD}\)+ \(\widehat{AOC}\)= 180o ( kề bù )
\(\rightarrow\)\(\widehat{AOD}\)= 180o - \(\widehat{AOC}\)= 180o - 125o = 55o
Có \(\widehat{BOD}\)+ \(\widehat{BOC}\)= 180o ( kề bù )
\(\rightarrow\)\(\widehat{BOD}\)= 180o - \(\widehat{BOC}\)
180o - 55o = 125o
Có : \(\widehat{AOC}=\widehat{BOD}\)( Hai góc đối đỉnh )
mà \(\widehat{AOC}=\widehat{BOD}\Rightarrow\widehat{BOD}=70^o\)
Có : \(\widehat{BOD}\)và \(\widehat{AOD}\)là hai góc kề bù
\(\Rightarrow\widehat{BOD}+\widehat{AOD}=180^o\)
\(70^o+\widehat{AOD}=180^o\Rightarrow\widehat{AOD}=180^o-70^o=110^o\)
Do \(\widehat{BOC}\)và \(\widehat{AOD}\)là hai góc đối đỉnh
\(\Rightarrow\widehat{BOC}=\widehat{AOD}\)
MÀ \(\widehat{AOD}=110^o\Rightarrow\widehat{BOC}=110^o\)